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Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It [#permalink]
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09 Feb 2011, 09:14
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Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It continues in such a way that by adding 5 to the nth term, one obtains the (n + 2)th term. What is the sum of the first 19 members of this sequence? A. 878 B. 900 C. 788 D. 928 E. 1022
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Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It [#permalink]
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09 Feb 2011, 10:12
I would split them up like this. 23 28 33 and so on (10 terms)....the 10th term = 23+9*5 = 68 27 32 37 and so on (9 terms)......the 9th term = 27+8*5 = 67 Since the distance between any two numbers is the same we can use arithmetics first+last/2 times no of numbers = the sum (23+68)/2 * 10 = 455 (27+67)/2 * 9 = 423 = 878
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Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It [#permalink]
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09 Feb 2011, 16:32
Thanks Mackieman! I did in the same way but got screwed as i included everything into the same formula n/2 (2a+ (n1) d )... it's the same approach
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Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It [#permalink]
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09 Feb 2011, 16:37
bhandariavi wrote: Thanks Mackieman! I did in the same way but got screwed as i included everything into the same formula n/2 (2a+ (n1) d )... it's the same approach Ah okay . Propably some of the Math Gurus know if there is a simpler and faster approach. "My" method shouldn't take more than 12 minutes though.
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Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It [#permalink]
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Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It
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